Consider the function f:RR defined by f(x)=cases(2-1x)|x|, & x00, & x=0cases.

Question

Consider the function $f:\mathbb{R}	o\mathbb{R}$ defined by $f(x)=\begin{cases}\left(2-\sin\frac1xight)|x|, & x
e0\0, & x=0\end{cases}$. Then, $f$ is

Accepted Answer

For $x>0$, $f(x)=\left(2-\sin\frac1x\right)x$, and near $x=\frac1\pi$ vs $x=\frac2\pi$ the oscillation of $\sin\frac1x$ causes $f$ to fail monotonicity: $f\left(\frac1\pi\right)=f\left(\frac2\pi\right)=\frac2\pi$, so $f$ is not monotonic on $(0,\infty)$. Similarly for $x<0$. Hence $f$ is non-monotonic on both $(-\infty,0)$ and $(0,\infty)$.

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